High Energy Physics (PHYS 557): solution to HW3 1/26/17 Exercises: • Show that γ5 = iγ0γ1γ2γ3 is hermitian and anticommutes with all the γµ. Show that in the Dirac-Pauli representation it takes the form (in 2 × 2 block notation) 0 1 γ5 = : 1 0 Show that, in the limit m ! 0, the solutions of the Dirac equation with definite helicity ~σ·~p p χ± p χ∓ u(p) = E + m and v(p) = E + m Ep+m ± p ~σ·~p χ ± p Ep+m ± χ∓ are eigenvectors of γ5 and show that the eigenvalues (which are called their \chirali- ties") have the same signs as their helicities for particles, and the opposite signs for antiparticles. Recall that the helicity spin-1/2 wavefuctions satisfy (note that p^ and not ~p appears on the left-hand side) ~σ · p^ χ± = ±χ± : We recall that γ0 is hermitian while ~γ is antihermitian. Thus γ5y = −i(−γ3)(−γ2)(−γ1)(γ0) = +iγ3γ2γ1γ0 = −iγ2γ1γ0γ3 = −iγ1γ0γ2γ3 = +iγ0γ1γ2γ3 = γ5 The explicit form is 1 0 0 σ 0 σ 0 σ γ5 = i 1 2 3 0 −1 −σ1 0 −σ2 0 −σ3 0 1 0 −iσ 0 0 σ = i 3 3 0 −1 0 −iσ3 −σ3 0 1 0 0 1 0 1 = = : 0 −1 −1 0 1 0 In the limit m ! 0, the spinors become p χ± p ∓χ∓ u(p)± = Ep and v(p)± = Ep : ±χ± χ∓ 5 These are manifestly eigenvectors of γ with chiralities ±1 for u(p)± and ∓1 for v(p)±. 1 Problems: 1. Helicity spin-1/2 wavefunctions. The defining equation for these wavefunctions is ~σ · p^ χ(^p)± = ±χ(^p)± ; (1) where now it has been made explicit that these wavefunctions depend on p^. We know that (making conventional choices of overall phases) 1 0 χ(^z) = and χ(^z) = : + 0 − 1 (a) Show that χ(^p)± = Rz(φ)Ry(θ)χ(^z)± ; (2) satisfy the helicity equation (1). Here the rotation matrix about an axis n^ by angle δ is Rn(δ) = exp(−iδn^ · ~σ=2) = cos(δ=2)1 − i sin(δ=2)^n · ~σ: Here θ and φ are, respectively, the polar and azimuthal angles of p^ relative to the coordinate axes. You can do this by explicit evaluation or using some of the rotation methodology you learned in graduate QM. Method 1: using spinology. The key ingredient is the transformation of spin-1/2 operators transform under rotations: ~σ · p^ = Rz(φ)Ry(θ)~σ · zR^ y(−θ)Rz(−φ) : See, for example, Sakurai, Modern QM, Eq. (3.2.47). Applying this χ(^p)± as given in Eq. (2), we find ~σ · pχ^ (^p)± = Rz(φ)Ry(θ)~σ · zR^ y(−θ)Rz(−φ)Rz(−φ)Ry(−θ)χ(^z)± = Rz(−phi)Ry(θ)~σ · zχ^ (^z)± = ±Rz(φ)Ry(θ)χ(^z)± = ±χ(^p)± : Method 2: direct evaluation. First we write out the claimed form in detail −iφ/2 e 0 c¯θ −s¯θ χ(^p)± = iφ/2 χ(^z)± 0 e s¯θ c¯θ wherec ¯θ ≡ cos(θ=2) ands ¯θ ≡ sin(θ=2). Evaluating the matrix products we find −iφ/2 −iφ e c¯θ iφ/2 e χ(^p)+ = iφ/2 = e c¯θ (3) e s¯θ t¯θ −iφ/2 −iφ −e s¯θ iφ/2 e χ(^p)− = iφ/2 = −e s¯θ ; (4) e c¯θ −1=t¯θ 2 −iφ where t¯θ = tan(θ=2). Now we need to express t¯θ and e in terms of the components ofp ^. First we use 1 − c 1 − p^ p2 (1 − p^ )2 ¯2 θ 3 12 3 cθ =p ^3 ) tθ = = = 2 = 2 ; 1 + cθ 1 +p ^3 (1 +p ^3) p12 2 2 2 2 ¯ where p12 =p ^1 +p ^2 = 1 − p^3. Since 0 ≤ θ=2 ≤ π=2 we know that tθ ≥ 0 so we should take the positive square root p12 1 − p^3 t¯θ = = : 1 +p ^3 p12 From the definition of φ we have p^ p^ p^ − ip^ e−iφ = cos φ − i sin φ = 1 − i 2 = 1 2 : p12 p12 p12 Combining all the above we find that p^1 − ip^2 p^1 − ip^2 χ(^p)+ / and χ(^p)− / : 1 − p^3 −(1 +p ^3) Finally we write out ~σ · p^ explicitly p^ p^ − ip^ ~σ · p^ = 3 1 2 ; p^1 + ip^2 −p^3 from which it is easy to check that the eigenvalues of χ(^p)± are ±1. (b) Evaluate y χ(^p)±χ(^z)±0 for each of the four possible choices of helicities. Using the results (3) and (4) from above, plus the forms of χ(^z)±, we find y iφ/2 χ(^p)+χ(^z)+ = e c¯θ ; y iφ/2 χ(^p)−χ(^z)+ = −e s¯θ ; y −iφ/2 χ(^p)+χ(^z)− = e s¯θ ; y −iφ/2 χ(^p)−χ(^z)− = e c¯θ : These results were used to calculate the Yukawa scattering matrix elements in class. (c) Show that the spin-averaged outer products of spinor solutions satisfy U ≡ u(p) u¯(p) + u(p) u¯(p) = p + m αβ +,α +,β −,α −,β = αβ V ≡ v(p) v¯(p) + v(p) v¯(p) = p − m : αβ +,α +,β −,α −,β = αβ 3 Here α and β are spinor indices that run from 1 − 4. These results are used in evaluating spin-averaged cross sections. Recall that ! j~pj ! p χ(^p)± p ∓ χ(^p)∓ u(p) = E + m and v(p) = E + m Ep+m : ± p ± j~pj χ(^p) ± p Ep+m ± χ(^p)∓ Evaluating the outer product for particle spinors, and recalling thatu ¯ = uyγ0, we find ! (Ep + m)M+ −|~pjM− U = 2 ; (5) j~pjM − ~p M − (Ep+m) + where the spin-wavefunction outer products are y y M± = χ(^p)+χ(^p)+ ± χ(^p)−χ(^p)− : We evaluate M+ using Eq. (2) together with y y χ(^p)± = χ(^z)±Ry(−θ)Rz(−φ) and y y χ(^z)+χ(^z)+ + χ(^z)−χ(^z)− = 1 ; finding M+ = Rz(−φ)Ry(−θ)1Ry(θ)Rz(φ) = 1 : To evaluate M− we use ~σ · p^M− = M+ = 1 )M− = ~σ · p^ ; 2 since (~σ · p^) = 1. Inserting the results for M± into Eq. (5) we find E + m −~σ · ~p U = p = p= + m : ~σ · ~p −Ep + m Now we repeat the exercise for the antiparticle spinors. Using their explicit form we find (E − m)M −|~pjM V = p + − j~pjM− −(Ep + m)M+ E − m −~σ · ~p = p ~σ · ~p −Ep − m = p= − m : 2. Dirac traceology. In class we showed, using only the defining anticommutation prop- erties of gamma matrices and the properties of γ5, that the trace of any odd number of gamma matrices vanishes and that tr(p==q) = 4p · q : Here we extend these results. 4 (a) Show that tr γ0n0 γ1n1 γ2n2 γ3n3 vanishes unless the positive integers n0, n1, n2 and n3 are all even. Note that this is a stronger than the result quoted above that the trace of an odd number of gamma matrices must vanish. It must be that the power of each of the gamma matrices separately must be even. Furthermore, since the different gamma ma- trices anticommute, this statement holds independent of the order of various gamma matrices. µ 2 Since (γ ) = ±1, we need only consider the cases nµ = 0 or 1. Since the trace of any odd number of gamma matrices vanishes, this leaves only the following cases to consider tr(γµγν) [with µ 6= ν] and tr(γ0γ1γ2γ3) / tr(γ5) : The former vanishes using the cyclicity of the trace and the defining property of gamma matrices 1 tr(γµγν) = tr(γνγµ) = tr [γµ; γν] = 4gµν = 0 for µ 6= ν : 2 + The latter vanishes using the fact that γ5 anticommutes with all the γµ: tr(γ5) = tr(γ0γ0γ5) = −tr(γ0γ5γ0) = −tr(γ5) = 0 : (b) Evaluate (providing explanation for your work) 5 tr(p=1p=2p=3p=4) and tr(γ p=1p=2p=3p=4) : To evaluate the first trace we move the p=1 through the trace until we return to the initial position. We need p=1p=2 = −p=2p=1 + 2p1 · p21 ; which follows from the defining anticommutation relation of gamma matrices. Thus we have tr(p=1p=2p=3p=4) = −tr(p=2p=1p=3p=4) + 8p1 · p2p3 · p4 = tr(p=2p=3p=1p=4) + 8p1 · p2p3 · p4 − 8p1 · p3p2 · p4 = −tr(p=1p=2p=3p=4) + 8p1 · p2p3 · p4 − 8p1 · p3p2 · p4 + 8p1 · p4p2 · p3 : Here at each stage we have used tr(p=1p=2) = 4p1 · p2. Now we have the same trace on both sides, but with opposite signs, so we can recombine these terms and divede by two leading to tr(p=1p=2p=3p=4) = 4p1 · p2p3 · p4 − 4p1 · p3p2 · p4 + 4p1 · p4p2 · p3 : 5 Now we turn to 5 5 µ ν ρ σ tr(γ p=1p=2p=3p=4) = tr(γ γ γ γ γ )p1,µp2,νp3,ρp4,σ : Since γ5 = iiγ0γ1γ2γ3 we know from the result of the first part of this problem that the trace is nonzero only when the indices µ, ν, ρ and σ are all different. If fµ, ν; ρ, σg = f0; 1; 2; 3g then the trace is −4i, since tr(γ5 2) = 4. If we permute the indices, then the anticommutation property of the gamma matrices will lead to a sign corresponding to the signature of the permutation, e.g. f1; 0; 2; 3g has signature −1. This totally antisymmetric property is represented by the four-index tensor.